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Question:

$\frac{1}{x^{4}-1}$

Solution:

$\frac{1}{\left(x^{4}-1\right)}=\frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)}=\frac{1}{(x+1)(x-1)\left(1+x^{2}\right)}$

Let $\frac{1}{(x+1)(x-1)\left(1+x^{2}\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^{2}+1\right)}$

$1=A(x-1)\left(x^{2}+1\right)+B(x+1)\left(x^{2}+1\right)+(C x+D)\left(x^{2}-1\right)$

$1=A\left(x^{3}+x-x^{2}-1\right)+B\left(x^{3}+x+x^{2}+1\right)+C x^{3}+D x^{2}-C x-D$

$1=(A+B+C) x^{3}+(-A+B+D) x^{2}+(A+B-C) x+(-A+B-D)$

Equating the coefficient of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+B+C=0$

$-A+B+D=0$

$A+B-C=0$

$-A+B-D=1$

On solving these equations, we obtain

$A=-\frac{1}{4}, B=\frac{1}{4}, C=0$, and $D=-\frac{1}{2}$

$\therefore \frac{1}{x^{4}-1}=\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2\left(x^{2}+1\right)}$

$\Rightarrow \int \frac{1}{x^{4}-1} d x=-\frac{1}{4} \log |x-1|+\frac{1}{4} \log |x-1|-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$

$\quad=\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$

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